Complex Numbers and Trigonometry

Date: 12/25/96 at 09:51:24
From: Thomas Glen Smith
Subject: Why doesn't asin(sin(s)) = s for complex numbers?
I calculate sin(3+4i) to be 3.85-27i, but when I compute
arcsin(3.85-27i), I get .14-4i for an answer. Why don't I get 3+4i?
I should think arcsin(sin(s)) should always produce s. It seems to
work for arcsin(sin(.14-4i)).

Date: 12/28/96 at 14:53:24
From: Doctor Pete
Subject: Re: Why doesn't asin(sin(s)) = s for complex numbers?
Hi,
The underlying reason why this phenomenon occurs is because inverse
trigonometric functions are not one-to-one when analytically continued
to the complex plane. This also explains, for instance, why there are
n n(th) roots of unity in the complex plane.
To see what's going on in further detail, let's first look at a
simpler example, the complex exponential. Euler's formula gives:
Exp[i*t] = Cos[t]+i*Sin[t]
where i is the imaginary unit Sqrt[-1]. Then the exponential of a
general complex number in rectangular form z = a+i*b is:
Exp[z] = Exp[a](Cos[b]+i*Sin[b])
From this, one can derive the inverse, the complex logarithm:
Log[z] = Ln[Abs[z]]+i*(Arg[z]+2*Pi*k), k = ...-2,-1,0,1,2,...
where Abs[z] is the magnitude of z, and Arg[z] is its *principal*
argument, the angle it forms with the origin when plotted as a vector
in the complex plane, restricted to some range, usually (-Pi, Pi].
Note that we have included a factor of 2*Pi*k. The reason for this is
because there are infinitely many ways of measuring the argument; for
instance,
Arg[1+i] = Pi/4
but
Arg[1+i] = {Pi/4, -7Pi/4, 9Pi/4, 17Pi/4, ... }.
Therefore, while we see that the real part of Log[z] is single-valued,
the imaginary part is multi-valued depending on what value of k we
choose. Indeed, this is the dilemma at hand, for is there any
particular reason why you should choose one k over another?
We see then, that Exp[z], compared to Log[z], is a fairly simple
function: for a given value of z, Exp[z] is unique, whereas Log[z] is
actually an infinite set of values. But this is deceptive, for in
reality, Exp[z] is just as complex (no pun intended) as Log[z]. A
little thought shows that:
Exp[a+i*b] = Exp[a+i*(b+2*Pi)] = Exp[a+i*(b+4*Pi)] = ...
In other words, there are infinitely many values of z which have the
same value of Exp[z]. So the full picture boils down to the fact that
Exp[z] is a many-to-one function, and Log[z] is a one-to-many
function.
Furthermore, this strange behavior owes itself to the simple fact that
Sin[x] and Cos[x] (for real x) are periodic functions with period
2*Pi.
Now, it should not be too surprising, then, that we can use Euler's
formula to obtain Sin[z] and Cos[z] for complex z. We have
Exp[i*z] = Cos[z] + i*Sin[z]
Exp[-i*z] = Cos[-z] + i*Sin[-z] = Cos[z] - i*Sin[z]
Adding these two gives:
2*Cos[z] = Exp[i*z] + Exp[-i*z]
Cos[z] = (Exp[i*z] + Exp[-i*z])/2
and subtracting gives:
2*i*Sin[z] = Exp[i*z] - Exp[-i*z]
Sin[z] = (Exp[i*z] - Exp[-i*z])/(2*i)
So the complex sine and cosine are expressible as sums of complex
exponentials, and therefore are many-to-one functions. Therefore
their inverses will behave like the complex logarithm in that they
will be one-to-many.
With this in mind, it is quite easy now to see why ArcSin[Sin[z]] is
not necessarily z. For if z1 and z2 are unequal and such that
Sin[z1] = Sin[z2], then ArcSin[Sin[z1]] is an infinite set which
contains z1 and z2. Furthermore, ArcSin[Sin[z2]] is that same
infinite set. Depending on which k-value you pick, you could even
have ArcSin[Sin[z1]] = z2 and vice versa. However, if you knew what
z1 was, then you could certainly pick ArcSin[Sin[z1]] = z1. But most
calculators are not sophisticated enough to consider this fact, and
instead simply choose the principal value (k=0) at all times.
It may seem strange at first, but these phenomena are all rooted in
the periodicity of the trigonometric functions. For example,
ArcSin[Sin[x]] = ArcSin[Sin[x + 2*Pi]]. The ideas are the same, only
with an added twist.
-Doctor Pete, The Math Forum
Check out our web site! http://mathforum.org/dr.math/